Let a and b be the vectors of the same magnitude such that ∣a + b| / ∣a - b∣ = √2 + 1. Then ∣a + b| + ∣a - b∣ / ∣a + b| - ∣a - b∣ is:

 

Question:

Let a and b be vectors of the same magnitude such that

$$\frac{\mid \mathbf{a}+\mathbf{b}\mid }{\mid \mathbf{a}-\mathbf{b}\mid }=\sqrt{2}+1.$$

Then the value of

$$\frac{\mid \mathbf{a}+\mathbf{b}\mid +\mid \mathbf{a}-\mathbf{b}\mid }{\mid \mathbf{a}+\mathbf{b}\mid -\mid \mathbf{a}-\mathbf{b}\mid }$$

is:

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Short Solution (Text):

Let

$$k=\frac{\mid \mathbf{a}+\mathbf{b}\mid }{\mid \mathbf{a}-\mathbf{b}\mid }=\sqrt{2}+1.$$

Divide numerator and denominator of the required expression by $|\mathbf{a}-\mathbf{b}|$:

$$\frac{\mid \mathbf{a}+\mathbf{b}\mid +\mid \mathbf{a}-\mathbf{b}\mid }{\mid \mathbf{a}+\mathbf{b}\mid -\mid \mathbf{a}-\mathbf{b}\mid }=\frac{k+1}{k-1}\mathrm{.}$$

Substitute $k=\sqrt{2}+1$:

$$\frac{k+1}{k-1}=\frac{(\sqrt{2}+1)+1}{(\sqrt{2}+1)-1}=\frac{\sqrt{2}+2}{\sqrt{2}}=1+\frac{2}{\sqrt{2}}=1+\sqrt{2}\mathrm{.<span\; class="mord\; sqrt"><span\; class="vlist-t\; vlist-t2"><span\; class="vlist-r"><span\; class="vlist-s"></span></span><span\; class="vlist-r"><span\; class="vlist"></span></span></span></span><span\; class="mord">.</span>}$$

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✅ Final Answer:

$$\boxed{1+\sqrt{2}}$$

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